Question

In the adjoining figure, side $BC$ of $△ABC$ is produced to $D$. The bisector of $\angle A$ meets $BC$ at $E$. Show that
$\angle ABC+\angle ACD=\angle AEC$.

Answer 1

Considering the question to be:

Show that $\angle ABC+\angle ACD=2\angle AEC$

Consider $\angle B=\alpha ,\angle BAE=\angle EAC=x$

Using exterior angle property, we have,

$\angle AEC=x+\alpha$

Similarly

$\angle ACD=2x+\alpha$

$\therefore \angle ABC+\angle ACD$

$=\alpha +(2x+\alpha )$

$=2x+2\alpha$

$=2(x+\alpha )$

$=2\angle AEC$

$\therefore \angle ABC+\angle ACD=2\angle AEC$